Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 6}{-6r - 12} \times \dfrac{-6r - 12}{r^2 - 14r + 48} $
Solution: First factor the quadratic. $q = \dfrac{r - 6}{-6r - 12} \times \dfrac{-6r - 12}{(r - 6)(r - 8)} $ Then factor out any other terms. $q = \dfrac{r - 6}{-6(r + 2)} \times \dfrac{-6(r + 2)}{(r - 6)(r - 8)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 6) \times -6(r + 2) } { -6(r + 2) \times (r - 6)(r - 8) } $ $q = \dfrac{ -6(r - 6)(r + 2)}{ -6(r + 2)(r - 6)(r - 8)} $ Notice that $(r + 2)$ and $(r - 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -6\cancel{(r - 6)}(r + 2)}{ -6(r + 2)\cancel{(r - 6)}(r - 8)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $q = \dfrac{ -6\cancel{(r - 6)}\cancel{(r + 2)}}{ -6\cancel{(r + 2)}\cancel{(r - 6)}(r - 8)} $ We are dividing by $r + 2$ , so $r + 2 \neq 0$ Therefore, $r \neq -2$ $q = \dfrac{-6}{-6(r - 8)} $ $q = \dfrac{1}{r - 8} ; \space r \neq 6 ; \space r \neq -2 $